Spectrum of operator on $L^2([0,1])$

Notice that the range of $T$ is the span of $\{1,x\}$, that is is the space of linear functions on $[0,1]$. So if $f(y)=ay+b$ is an eigenvector of $T$ corresponding to $\lambda\in\mathbb{C}\setminus\{0\}$, $$ (\frac12 a+ b)x + (\frac13 a + \frac12b) = \lambda a x + \lambda b $$ Thus $$ \begin{pmatrix} \frac12-\lambda & 1\\ \frac13 & \frac12-\lambda \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix} $$

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Functional Analysis

Spectral Theory

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